vanishing fourier series

Question

Let $f(x)=\sum\limits_{k\geq 2} a_k \cos(kx)+b_k\sin(kx)$ a Fourier series of a real-valued continuous function $f$ (with $2\pi$-periodicity).Note here that $f$ is orthogonal to $1,\cos$ and $\sin$ on $[-\pi,\pi]$. Is it true that $f$ vanishes on $[0,\pi]$ ?

Indeed, that is a particular case $N=1$ of a more general question : if the Fourier series of a real-valued function $f$ contains no terms in $\cos(kx)$ or $\sin(kx)$ with $|k|< N$, is it true that $f$ has a zero on any interval of lengh $2\pi/N$ ?

Answer 1

So $f(x)=\cos(2 x)$ sounds valid, and $f(\pi/4)=0$.

Answer 2

If $a_k=b_k =0$ Except for $a_2=1$ then F(x)=cos(2x) which certainly does not vanish on all of [0,$\pi$] . If you mean vanish at some point ,I don't know .

Answer 3

If we partition the $[-\pi,\pi]$ into 4 intervals, where $[0,\pi]$ is a sub interval of one of partitioned intervals... say

$[-\pi, -\frac {2\pi}{3}), [-\frac {2\pi}{3},-\frac {\pi}{3}),[-\frac {\pi}{3},0),[0,\pi]$

Suppose is constant in each of the 4 intervals

i.e. $f(x) = \begin {cases}a &x\in[-\pi, -\frac {2\pi}{3})\\b&x\in[-\frac {2\pi}{3},-\frac {\pi}{3})\\c&x\in[-\frac {\pi}{3},0)\\d&x\in[0,\pi]\end{cases}$

We must find $a,b,c,d$ such that.

$\int_{-\pi}^{\pi} f(x) = 0\\ \int_{-\pi}^{\pi} f(x)\sin x = 0\\ \int_{-\pi}^{\pi} f(x)\cos x = 0$

That will be system of linear equations and will have a non-trivial kernel.

This is not a continuous function, and it does cross $0$ at the endpoints of the interval $[0,\pi]$. Nonetheless, we can use this approach to structure a function that is continuous and has parameters such that

$\int_{-\pi}^{\pi} f(x) = 0\\ \int_{-\pi}^{\pi} f(x)\sin x = 0\\ \int_{-\pi}^{\pi} f(x)\cos x = 0$

Produce the necessary set of linear equations.